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MCQ Questions - Topic-wise
Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


Case Study / Scenario-Based MCQs for Sub-Topics of Topic 7: Mensuration
Content On This Page
Introduction to Mensuration: Perimeter and Area Concepts Perimeter of Various Plane Figures Area of Basic Plane Figures
Area of Triangles using Heron's Formula Area of Quadrilaterals and Polygons (Advanced) Circles: Circumference and Area
Areas Related to Circles: Sectors and Segments Area and Perimeter of Combinations of Plane Figures Introduction to Solid Shapes and Basic Terms
Surface Area of Standard Solids Volume of Standard Solids Surface Area and Volume of Combined Solids
Conversion of Solids and Frustum of a Cone


Case Study / Scenario-Based MCQs for Sub-Topics of Topic 7: Mensuration



Introduction to Mensuration: Perimeter and Area Concepts

Question 1. Ramesh wants to put a decorative lace around a rectangular table cloth that is 1.5 m long and 1 m wide. He also wants to buy a plastic sheet to cover the top of the table cloth.

Which mensuration concepts should Ramesh use to determine the length of lace needed and the amount of plastic sheet needed, respectively?

(A) Area for lace, Perimeter for sheet

(B) Volume for lace, Area for sheet

(C) Perimeter for lace, Area for sheet

(D) Perimeter for lace, Volume for sheet

Answer:

Question 2. A farmer owns a square field in his village. He wants to fence the entire boundary of the field and also needs to estimate the amount of fertilizer required for the entire field area.

If the side of the square field is $100\ \text{m}$, what is the length of the fence required and the area of the field in square metres?

(A) Fence: $100\ \text{m}$, Area: $10000\ \text{m}^2$

(B) Fence: $400\ \text{m}$, Area: $10000\ \text{m}^2$

(C) Fence: $400\ \text{m}$, Area: $400\ \text{m}^2$

(D) Fence: $10000\ \text{m}^2$, Area: $400\ \text{m}$

Answer:

Question 3. A painter is hired to paint the four walls of a rectangular room. The room is 5 m long, 4 m wide, and 3 m high. He needs to calculate the total area of the walls to determine the amount of paint.

Which concept is most directly relevant to calculate the paintable area?

(A) Perimeter of the floor

(B) Area of the floor

(C) Surface area of the cuboid (excluding floor and ceiling)

(D) Volume of the room

Answer:

Question 4. A constructor is building a cylindrical water tank. He needs to know how much water it can hold and how much material is needed to build its curved side.

Which mensuration concepts correspond to the water capacity and the material for the side, respectively?

(A) Volume, Surface Area

(B) Surface Area, Volume

(C) Perimeter, Area

(D) Area, Perimeter

Answer:

Question 5. Look at the units: $\text{m}^2$, $\text{cm}$, $\text{L}$, $\text{km}^3$.

Which unit is appropriate for measuring the area of a small classroom floor?

(A) $\text{m}^2$

(B) $\text{cm}$

(C) $\text{L}$

(D) $\text{km}^3$

Answer:

Question 6. A farmer has a rectangular plot of land $500\ \text{m}$ long and $200\ \text{m}$ wide. He wants to divide it into smaller rectangular plots, each $100\ \text{m}$ long and $50\ \text{m}$ wide.

To determine how many smaller plots he can make, he needs to use which mensuration concept for both the large and small plots?

(A) Perimeter

(B) Area

(C) Volume

(D) Length

Answer:

Question 7. A student is drawing a map of a city park which is triangular in shape. He needs to draw the boundary and indicate the total covered green space.

Which mensuration concepts should he use for the boundary and the green space, respectively?

(A) Area, Perimeter

(B) Perimeter, Area

(C) Length, Width

(D) Volume, Surface Area

Answer:

Question 8. A carpenter is making a wooden box. He needs to find the amount of wood required to build the entire box and the amount of space inside the box to store things.

Which concepts correspond to the wood material and the storage space, respectively?

(A) Volume, Surface Area

(B) Perimeter, Area

(C) Surface Area, Volume

(D) Area, Perimeter

Answer:

Question 9. A tailor is stitching a border around a circular tablecloth. The radius of the tablecloth is $0.7\ \text{m}$.

What measurement does the tailor need to calculate to buy the correct length of border material?

(A) Area of the tablecloth

(B) Diameter of the tablecloth

(C) Circumference of the tablecloth

(D) Radius of the tablecloth

Answer:

Question 10. An architect is designing a building. He needs to calculate the total built-up area on each floor and the total external surface area of the building for planning and costing purposes.

What concepts are he primarily using for these calculations?

(A) Perimeter for both

(B) Area for built-up area, Surface Area for external surface

(C) Volume for both

(D) Surface Area for built-up area, Area for external surface

Answer:



Perimeter of Various Plane Figures

Question 1. A school playground is rectangular, 60 m long and 40 m wide. The school wants to put a boundary line around the playground for sports activities.

What is the total length of the boundary line required?

(A) $200\ \text{m}$

(B) $100\ \text{m}$

(C) $2400\ \text{m}$

(D) $120\ \text{m}$

Answer:

Question 2. A farmer has a triangular field with side lengths $15\ \text{m}$, $20\ \text{m}$, and $25\ \text{m}$. He plans to fence it with 3 rounds of wire.

What is the total length of wire needed?

(A) $60\ \text{m}$

(B) $120\ \text{m}$

(C) $180\ \text{m}$

(D) $300\ \text{m}$

Answer:

Question 3. A wall in a living room is rectangular with a length of 8 m and a height of 3 m. A decorative border needs to be placed along the top and bottom edges of the wall.

What is the total length of the border required?

(A) $11\ \text{m}$

(B) $16\ \text{m}$

(C) $22\ \text{m}$

(D) $24\ \text{m}$

Answer:

Question 4. A picture frame is in the shape of a square. The perimeter of the frame is $120\ \text{cm}$.

What is the side length of the picture frame?

(A) $30\ \text{cm}$

(B) $40\ \text{cm}$

(C) $60\ \text{cm}$

(D) $120\ \text{cm}$

Answer:

Question 5. A tailor is making a border for a triangular scarf with sides $50\ \text{cm}$, $50\ \text{cm}$, and $80\ \text{cm}$.

What is the total length of the border needed for the scarf?

(A) $130\ \text{cm}$

(B) $180\ \text{cm}$

(C) $200\ \text{cm}$

(D) $260\ \text{cm}$

Answer:

Question 6. The boundary of a rectangular park is 400 m. If the length of the park is 120 m, what is its width?

(A) $80\ \text{m}$

(B) $100\ \text{m}$

(C) $140\ \text{m}$

(D) $280\ \text{m}$

Answer:

Question 7. A polygonal garden has sides measuring 10 m, 15 m, 12 m, 18 m, and 9 m. To put a fence around it, the length of the fence will be the sum of these side lengths.

What is the perimeter of the garden?

(A) $60\ \text{m}$

(B) $64\ \text{m}$

(C) $74\ \text{m}$

(D) $80\ \text{m}$

Answer:

Question 8. A runner completes 5 laps around a rectangular track that is 80 m long and 30 m wide.

What is the total distance covered by the runner?

(A) $110\ \text{m}$

(B) $220\ \text{m}$

(C) $1100\ \text{m}$

(D) $2200\ \text{m}$

Answer:

Question 9. The perimeter of a parallelogram is $90\ \text{cm}$. If one side is $25\ \text{cm}$, what is the length of the adjacent side?

(A) $20\ \text{cm}$

(B) $45\ \text{cm}$

(C) $65\ \text{cm}$

(D) $15\ \text{cm}$

Answer:

Question 10. A wire of length $16\ \text{cm}$ is bent into the shape of a square. What is the side length of the square?

If the same wire is then bent into a rectangle with length $5\ \text{cm}$, what is the width of the rectangle?

(A) Square side: $4\ \text{cm}$, Rectangle width: $3\ \text{cm}$

(B) Square side: $4\ \text{cm}$, Rectangle width: $11\ \text{cm}$

(C) Square side: $16\ \text{cm}$, Rectangle width: $3\ \text{cm}$

(D) Square side: $8\ \text{cm}$, Rectangle width: $3\ \text{cm}$

Answer:



Area of Basic Plane Figures

Question 1. A room floor is rectangular, 7 m long and 5 m wide. It needs to be tiled. Each square tile has a side length of 0.5 m.

What is the area of the floor to be tiled?

(A) $12\ \text{m}^2$

(B) $24\ \text{m}^2$

(C) $35\ \text{m}^2$

(D) $70\ \text{m}^2$

Answer:

Question 2. A parallelogram-shaped piece of glass has a base of $25\ \text{cm}$ and a corresponding height of $10\ \text{cm}$.

What is the area of the glass piece?

(A) $35\ \text{cm}^2$

(B) $70\ \text{cm}^2$

(C) $125\ \text{cm}^2$

(D) $250\ \text{cm}^2$

Answer:

Question 3. A triangular banner needs to be painted. Its base is $4\ \text{m}$ and its height is $3\ \text{m}$.

What is the area of the banner?

(A) $6\ \text{m}^2$

(B) $7\ \text{m}^2$

(C) $12\ \text{m}^2$

(D) $24\ \text{m}^2$

Answer:

Question 4. A square piece of paper has a side length of $15\ \text{cm}$. It is cut into smaller square pieces with side length $3\ \text{cm}$.

How many smaller square pieces can be cut from the larger square?

(A) 5

(B) 10

(C) 15

(D) 25

Answer:

Question 5. A rectangular field has an area of $480\ \text{m}^2$. If its length is $30\ \text{m}$, what is its width?

(A) $16\ \text{m}$

(B) $18\ \text{m}$

(C) $20\ \text{m}$

(D) $24\ \text{m}$

Answer:

Question 6. The cost of painting a wall at $\textsf{₹}\, 15$ per square metre is $\textsf{₹}\, 900$. If the length of the wall is $4\ \text{m}$, what is its height?

(A) $15\ \text{m}$

(B) $60\ \text{m}$

(C) $12\ \text{m}$

(D) $15/4 \approx 3.75\ \text{m}$

Answer:

Question 7. A parallelogram and a triangle have the same base of length $10\ \text{cm}$. If they are between the same parallel lines, and the height is $7\ \text{cm}$, what is the area of the triangle?

(A) $35\ \text{cm}^2$

(B) $70\ \text{cm}^2$

(C) $10\ \text{cm}^2$

(D) $17\ \text{cm}^2$

Answer:

Question 8. A plot of land is in the shape of a trapezium with parallel sides $15\ \text{m}$ and $25\ \text{m}$, and the perpendicular distance between them is $10\ \text{m}$.

What is the area of the trapezium?

(A) $40\ \text{m}^2$

(B) $200\ \text{m}^2$

(C) $400\ \text{m}^2$

(D) $800\ \text{m}^2$

Answer:

Question 9. A rectangular sheet of metal is $40\ \text{cm}$ by $25\ \text{cm}$. If 4 identical squares of side $2\ \text{cm}$ are cut from the four corners, what is the area of the remaining sheet?

(A) $1000\ \text{cm}^2$

(B) $984\ \text{cm}^2$

(C) $992\ \text{cm}^2$

(D) $968\ \text{cm}^2$

Answer:

Question 10. The area of a square field is $6400\ \text{m}^2$. A rectangular field has twice the length and half the width of the square field. What is the area of the rectangular field?

(A) $3200\ \text{m}^2$

(B) $6400\ \text{m}^2$

(C) $12800\ \text{m}^2$

(D) $25600\ \text{m}^2$

Answer:



Area of Triangles using Heron's Formula

Question 1. A farmer wants to find the area of his triangular field whose sides measure $40\ \text{m}$, $60\ \text{m}$, and $80\ \text{m}$.

What is the semi-perimeter of the field?

(A) $90\ \text{m}$

(B) $180\ \text{m}$

(C) $120\ \text{m}$

(D) $60\ \text{m}$

Answer:

Question 2. For the triangular field in Question 1, calculate its area using Heron's formula. Area $= \sqrt{s(s-a)(s-b)(s-c)}$.

(A) $\sqrt{90(50)(30)(10)}\ \text{m}^2$

(B) $\sqrt{90 \times 50 \times 30 \times 10} = \sqrt{9 \times 10 \times 5 \times 10 \times 3 \times 10 \times 10} = \sqrt{9 \times 5 \times 3 \times 10^4} = 3 \times \sqrt{15} \times 100 = 300\sqrt{15}\ \text{m}^2$

(C) $300\sqrt{15}\ \text{m}^2$

(D) Both (B) and (C) are correct.

Answer:

Question 3. An equilateral triangle-shaped garden has a side length of $20\ \text{m}$.

Using Heron's formula, what is its area? (Hint: $s=30$, $s-a=10, s-b=10, s-c=10$)

(A) $\sqrt{30 \times 10 \times 10 \times 10} = \sqrt{3 \times 10^4} = 100\sqrt{3}\ \text{m}^2$

(B) $200\sqrt{3}\ \text{m}^2$

(C) $100\sqrt{3}\ \text{m}^2$

(D) Both (A) and (C) are correct.

Answer:

Question 4. A school has a small triangular plot for a flower bed. The sides are 13 m, 14 m, and 15 m. To purchase soil for the bed, the area needs to be known.

What is the area of the flower bed using Heron's formula?

(A) $42\ \text{m}^2$

(B) $84\ \text{m}^2$

(C) $90\ \text{m}^2$

(D) $180\ \text{m}^2$

Answer:

Question 5. An isosceles triangle has equal sides of $10\ \text{cm}$ and a base of $12\ \text{cm}$.

What is the area of this triangle using Heron's formula? (Hint: $s=16$)

(A) $\sqrt{16(16-10)(16-10)(16-12)} = \sqrt{16 \times 6 \times 6 \times 4} = 4 \times 6 \times 2 = 48\ \text{cm}^2$

(B) $32\ \text{cm}^2$

(C) $48\ \text{cm}^2$

(D) Both (A) and (C) are correct.

Answer:

Question 6. A triangular piece of land has a perimeter of $120\ \text{m}$. Two of its sides are $30\ \text{m}$ and $40\ \text{m}$. The cost of planting grass at $\textsf{₹}\, 100$ per $10\ \text{m}^2$ is needed.

What is the length of the third side of the triangle?

(A) $50\ \text{m}$

(B) $60\ \text{m}$

(C) $70\ \text{m}$

(D) $80\ \text{m}$

Answer:

Question 7. For the triangular land in Question 6, calculate its area using Heron's formula. (Sides are 30, 40, 50).

(A) $600\ \text{m}^2$

(B) $1200\ \text{m}^2$

(C) $1500\ \text{m}^2$

(D) $2400\ \text{m}^2$

Answer:

Question 8. For the triangular land in Question 6, what is the total cost of planting grass?

(A) $\textsf{₹}\, 6000$

(B) $\textsf{₹}\, 60000$

(C) $\textsf{₹}\, 12000$

(D) $\textsf{₹}\, 120000$

Answer:

Question 9. The sides of a triangle are in the ratio $12:17:25$ and its perimeter is $540\ \text{cm}$. To find its area, first find the actual side lengths.

What are the side lengths of the triangle?

(A) 120 cm, 170 cm, 250 cm

(B) 100 cm, 150 cm, 290 cm

(C) 12 cm, 17 cm, 25 cm

(D) 24 cm, 34 cm, 50 cm

Answer:

Question 10. For the triangle in Question 9, calculate its area using Heron's formula. (Sides 120, 170, 250)

(A) $9000\ \text{cm}^2$

(B) $18000\ \text{cm}^2$

(C) $36000\ \text{cm}^2$

(D) $4500\ \text{cm}^2$

Answer:



Area of Quadrilaterals and Polygons (Advanced)

Question 1. A park is in the shape of a quadrilateral ABCD. A diagonal AC is $24\ \text{m}$ long. The perpendiculars from B and D to AC are $8\ \text{m}$ and $13\ \text{m}$ respectively.

What is the area of the park?

(A) $12 \times (8+13)\ \text{m}^2$

(B) $24 \times (8+13)\ \text{m}^2$

(C) $12 \times 8 + 12 \times 13\ \text{m}^2$

(D) Both (A) and (C) are correct.

Answer:

Question 2. A field is in the shape of a rhombus. The lengths of its diagonals are $30\ \text{m}$ and $16\ \text{m}$. The cost of cultivating the field at $\textsf{₹}\, 5$ per $\text{m}^2$ is needed.

What is the area of the field?

(A) $240\ \text{m}^2$

(B) $480\ \text{m}^2$

(C) $150\ \text{m}^2$

(D) $80\ \text{m}^2$

Answer:

Question 3. For the rhombus field in Question 2, what is the total cost of cultivating the field?

(A) $\textsf{₹}\, 1200$

(B) $\textsf{₹}\, 2400$

(C) $\textsf{₹}\, 750$

(D) $\textsf{₹}\, 400$

Answer:

Question 4. A piece of colored paper is in the shape of a kite. The diagonals measure $40\ \text{cm}$ and $25\ \text{cm}$.

What is the area of the colored paper?

(A) $1000\ \text{cm}^2$

(B) $500\ \text{cm}^2$

(C) $65\ \text{cm}^2$

(D) $130\ \text{cm}^2$

Answer:

Question 5. A farmer wants to find the area of his quadrilateral field ABCD. He measures the sides as AB=$10\ \text{m}$, BC=$17\ \text{m}$, CD=$15\ \text{m}$, DA=$18\ \text{m}$, and diagonal AC=$20\ \text{m}$.

To find the area, he can calculate the area of triangle ABC and triangle ADC and sum them. What is the semi-perimeter of triangle ABC?

(A) $20\ \text{m}$

(B) $23.5\ \text{m}$

(C) $28.5\ \text{m}$

(D) $47\ \text{m}$

Answer:

Question 6. For the quadrilateral field in Question 5, what is the area of triangle ABC (sides 10, 17, 20)? Use Heron's formula.

(A) $\sqrt{23.5(13.5)(6.5)(3.5)}$

(B) $\approx 84.9\ \text{m}^2$

(C) $120\ \text{m}^2$

(D) Both (A) and (B) are calculations/approximate results.

Answer:

Question 7. For the quadrilateral field in Question 5, what is the semi-perimeter of triangle ADC (sides 20, 15, 18)?

(A) $26.5\ \text{m}$

(B) $33.5\ \text{m}$

(C) $53\ \text{m}$

(D) $20\ \text{m}$

Answer:

Question 8. For the quadrilateral field in Question 5, calculate the area of triangle ADC (sides 20, 15, 18) using Heron's formula.

(A) $\sqrt{26.5(6.5)(11.5)(8.5)}$

(B) $\approx 114.3\ \text{m}^2$

(C) $90\ \text{m}^2$

(D) Both (A) and (B) are calculations/approximate results.

Answer:

Question 9. What is the total area of the quadrilateral field ABCD in Question 5?

(A) $84.9 + 114.3 = 199.2\ \text{m}^2$ (approx)

(B) $\approx 200\ \text{m}^2$ (approx)

(C) $220\ \text{m}^2$

(D) Both (A) and (B) are approximate results.

Answer:

Question 10. A plot of land is in the shape of a regular pentagon with a side length of $10\ \text{m}$. To find its area, one method is to divide it into 5 identical isosceles triangles meeting at the center. The distance from the center to the midpoint of a side (apothem) is approximately $6.88\ \text{m}$.

Using the apothem, what is the area of the regular pentagon?

(A) Area of one triangle $= \frac{1}{2} \times 10 \times 6.88\ \text{m}^2 = 34.4\ \text{m}^2$.

(B) Total area $= 5 \times 34.4\ \text{m}^2 = 172\ \text{m}^2$.

(C) Perimeter $= 5 \times 10 = 50\ \text{m}$. Area $= \frac{1}{2} \times \text{Perimeter} \times \text{Apothem} = \frac{1}{2} \times 50 \times 6.88 = 172\ \text{m}^2$.

(D) All of the above calculations and result are correct (approximate).

Answer:



Circles: Circumference and Area

Question 1. A circular clock face has a radius of $14\ \text{cm}$. A decorative metal strip is to be placed around its edge.

What is the length of the metal strip needed? (Use $\pi = \frac{22}{7}$)

(A) $22\ \text{cm}$

(B) $44\ \text{cm}$

(C) $88\ \text{cm}$

(D) $154\ \text{cm}$

Answer:

Question 2. A circular carpet needs to be placed in a room. The diameter of the carpet is $3.5\ \text{m}$.

What is the area of the carpet? (Use $\pi = \frac{22}{7}$)

(A) $9.625\ \text{m}^2$

(B) $12.25\ \text{m}^2$

(C) $38.5\ \text{m}^2$

(D) $10.5\ \text{m}^2$

Answer:

Question 3. A wheel of a cart has a radius of $42\ \text{cm}$. How much distance does the cart cover when the wheel makes 50 complete revolutions? (Use $\pi = \frac{22}{7}$)

(A) $2.64\ \text{m}$

(B) $264\ \text{m}$

(C) $132\ \text{m}$

(D) $1320\ \text{m}$

Answer:

Question 4. The circumference of a circular field is $352\ \text{m}$. What is the radius of the field? (Use $\pi = \frac{22}{7}$)

(A) $28\ \text{m}$

(B) $56\ \text{m}$

(C) $112\ \text{m}$

(D) $176\ \text{m}$

Answer:

Question 5. For the circular field in Question 4, what is the area of the field? (Use $\pi = \frac{22}{7}$)

(A) $2464\ \text{m}^2$

(B) $4928\ \text{m}^2$

(C) $9856\ \text{m}^2$

(D) $19712\ \text{m}^2$

Answer:

Question 6. A wire is bent into a circle of radius $7\ \text{cm}$. If it is rebent into a square, what is the side length of the square? (Use $\pi = \frac{22}{7}$)

(A) $7\ \text{cm}$

(B) $11\ \text{cm}$

(C) $14\ \text{cm}$

(D) $22\ \text{cm}$

Answer:

Question 7. The area of a circular garden is $1256\ \text{m}^2$. Using $\pi = 3.14$, what is the radius of the garden?

(A) $10\ \text{m}$

(B) $20\ \text{m}$

(C) $40\ \text{m}$

(D) $100\ \text{m}$

Answer:

Question 8. Two concentric circles have radii $10\ \text{cm}$ and $7\ \text{cm}$. What is the area of the ring between the two circles? (Use $\pi = \frac{22}{7}$)

(A) $110\ \text{cm}^2$

(B) $176\ \text{cm}^2$

(C) $220\ \text{cm}^2$

(D) $264\ \text{cm}^2$

Answer:

Question 9. A farmer irrigates a circular field using a central pivot system. The irrigation arm rotates and covers a circular area. If the arm is $20\ \text{m}$ long, what is the maximum area of the field that can be irrigated in one full rotation? (Use $\pi = 3.14$)

(A) $1256\ \text{m}^2$

(B) $628\ \text{m}^2$

(C) $400\ \text{m}^2$

(D) $12560\ \text{m}^2$

Answer:

Question 10. A bicycle wheel has a diameter of $70\ \text{cm}$. To travel 1 km, how many revolutions must the wheel make? (Use $\pi = \frac{22}{7}$ and convert units)

(A) Approx. 455 revolutions

(B) Approx. 910 revolutions

(C) Approx. 227 revolutions

(D) Approx. 1820 revolutions

Answer:



Areas Related to Circles: Sectors and Segments

Question 1. A pizza is cut into 8 equal slices. If the pizza has a radius of $14\ \text{cm}$, each slice is a sector of the circle.

What is the central angle of each slice?

(A) $45^\circ$

(B) $90^\circ$

(C) $60^\circ$

(D) $30^\circ$

Answer:

Question 2. For the pizza slice in Question 1 (radius $14\ \text{cm}$, angle $45^\circ$), what is the area of one slice? (Use $\pi = \frac{22}{7}$)

(A) $77\ \text{cm}^2$

(B) $154\ \text{cm}^2$

(C) $77/2\ \text{cm}^2$

(D) $308\ \text{cm}^2$

Answer:

Question 3. A chord of a circle of radius $10\ \text{cm}$ subtends a right angle at the centre. The area of the corresponding sector is needed. (Use $\pi = 3.14$).

What is the area of this sector?

(A) $314\ \text{cm}^2$

(B) $157\ \text{cm}^2$

(C) $78.5\ \text{cm}^2$

(D) $31.4\ \text{cm}^2$

Answer:

Question 4. For the circle and chord in Question 3 (radius $10\ \text{cm}$, angle $90^\circ$), what is the area of the triangle formed by the radii and the chord?

(A) $100\ \text{cm}^2$

(B) $50\ \text{cm}^2$

(C) $25\ \text{cm}^2$

(D) $75\ \text{cm}^2$

Answer:

Question 5. For the circle and chord in Question 3 (radius $10\ \text{cm}$, angle $90^\circ$), what is the area of the corresponding minor segment? (Use $\pi = 3.14$)

(A) $28.5\ \text{cm}^2$

(B) $50\ \text{cm}^2$

(C) $78.5\ \text{cm}^2$

(D) $128.5\ \text{cm}^2$

Answer:

Question 6. A gardener is designing a circular flower bed of radius $28\ \text{m}$. He plans to dedicate a sector with a central angle of $120^\circ$ to rose bushes. What is the area for rose bushes? (Use $\pi = \frac{22}{7}$)

(A) $616\ \text{m}^2$

(B) $1848\ \text{m}^2$

(C) $1232/3\ \text{m}^2$

(D) $1232\ \text{m}^2$

Answer:

Question 7. The minute hand of a clock is $10\ \text{cm}$ long. How much area does the minute hand sweep in 30 minutes? (Use $\pi = 3.14$)

(A) $31.4\ \text{cm}^2$

(B) $62.8\ \text{cm}^2$

(C) $157\ \text{cm}^2$

(D) $314\ \text{cm}^2$

Answer:

Question 8. A sector has radius $r$ and arc length $l$. Its area is $\frac{1}{2}rl$. If the arc length is $22\ \text{cm}$ and the radius is $7\ \text{cm}$, what is the area of the sector?

(A) $77\ \text{cm}^2$

(B) $154\ \text{cm}^2$

(C) $38.5\ \text{cm}^2$

(D) $11\ \text{cm}^2$

Answer:

Question 9. A chord of a circle is equal to the radius of the circle. What is the angle subtended by the chord at the centre?

(A) $30^\circ$

(B) $45^\circ$

(C) $60^\circ$

(D) $90^\circ$

Answer:

Question 10. For the scenario in Question 9 (chord = radius), if the radius is $7\ \text{cm}$, what is the area of the triangle formed by the radii and the chord? (Use $\sin 60^\circ = \frac{\sqrt{3}}{2}$)

(A) $49\sqrt{3}/2\ \text{cm}^2$

(B) $49/2\ \text{cm}^2$

(C) $49\sqrt{3}\ \text{cm}^2$

(D) $7\sqrt{3}/2\ \text{cm}^2$

Answer:



Area and Perimeter of Combinations of Plane Figures

Question 1. A shape is made by joining a square of side $10\ \text{cm}$ and a semicircle on one of its sides as a diameter. What is the area of the combined shape? (Use $\pi = 3.14$)

(A) $100 + \frac{1}{2}\pi (5)^2 = 100 + 12.5\pi \approx 100 + 39.25 = 139.25\ \text{cm}^2$

(B) $100 + \pi (10)^2 = 100 + 100\pi \approx 414\ \text{cm}^2$

(C) $100 + \frac{1}{2}\pi (10)^2 = 100 + 50\pi \approx 257\ \text{cm}^2$

(D) $100\ \text{cm}^2$

Answer:

Question 2. For the shape in Question 1 (square of side $10\ \text{cm}$ with semicircle on one side), what is the perimeter of the combined shape? (Use $\pi = 3.14$)

(A) $10+10+10 + \frac{1}{2}(2\pi \times 5) = 30 + 5\pi \approx 30 + 15.7 = 45.7\ \text{cm}$

(B) $4 \times 10 + 2\pi \times 5 = 40 + 10\pi \approx 71.4\ \text{cm}$

(C) $10+10+10+10 = 40\ \text{cm}$

(D) $30 + 10\pi \approx 61.4\ \text{cm}$

Answer:

Question 3. A rectangular park is $20\ \text{m}$ by $15\ \text{m}$. A path $2\ \text{m}$ wide runs inside it, along the boundary. What is the area of the path?

(A) Area of park $= 20 \times 15 = 300\ \text{m}^2$. Inner rectangle dimensions: $(20-2-2) \times (15-2-2) = 16 \times 11 = 176\ \text{m}^2$. Area of path $= 300 - 176 = 124\ \text{m}^2$.

(B) Area of park $= 300\ \text{m}^2$. Inner rectangle $18 \times 13 = 234\ \text{m}^2$. Area of path $= 300 - 234 = 66\ \text{m}^2$.

(C) Area of path $= 2 \times (20+15) \times 2 = 4 \times 35 = 140\ \text{m}^2$ (Incorrect method).

(D) $124\ \text{m}^2$

Answer:

Question 4. A design is made by drawing quadrants of circles at the four corners of a square plot of side $10\ \text{m}$. Each quadrant has a radius of $3.5\ \text{m}$. A circle is also drawn in the center with diameter $4\ \text{m}$. The area of the remaining portion of the square plot is needed. (Use $\pi = \frac{22}{7}$)

What is the area of the square plot?

(A) $100\ \text{m}^2$

(B) $400\ \text{m}^2$

(C) $10\ \text{m}^2$

(D) $50\ \text{m}^2$

Answer:

Question 5. For the design in Question 4, what is the total area of the four quadrants? (Radius $3.5\ \text{m}$)

(A) Area of one quadrant $= \frac{1}{4} \times \frac{22}{7} \times (3.5)^2 = \frac{1}{4} \times \frac{22}{7} \times 12.25 = \frac{1}{4} \times 22 \times 1.75 = 12.25\ \text{m}^2$. Total area $= 4 \times 12.25 = 49\ \text{m}^2$.

(B) $4 \times \frac{1}{4} \times \pi (3.5)^2 = \pi (3.5)^2 = \frac{22}{7} \times 12.25 = 38.5\ \text{m}^2$.

(C) $4 \times \frac{1}{2} \pi (3.5)^2 = 2\pi (3.5)^2 = 77\ \text{m}^2$.

(D) $38.5\ \text{m}^2$

Answer:

Question 6. For the design in Question 4, what is the area of the central circle? (Diameter $4\ \text{m}$)

(A) $\pi (4)^2 = 16\pi\ \text{m}^2 \approx 50.24\ \text{m}^2$

(B) $\pi (2)^2 = 4\pi\ \text{m}^2 \approx 12.56\ \text{m}^2$

(C) $\pi (1)^2 = \pi\ \text{m}^2 \approx 3.14\ \text{m}^2$

(D) $\pi (3.5)^2 \approx 38.5\ \text{m}^2$

Answer:

Question 7. For the design in Question 4, what is the area of the remaining portion of the square plot?

(A) Area of square $-$ (Area of 4 quadrants $+$ Area of circle)

(B) $100 - (38.5 + 12.56) = 100 - 51.06 = 48.94\ \text{m}^2$ (approx)

(C) $100 - 38.5 = 61.5\ \text{m}^2$

(D) $100\ \text{m}^2$

Answer:

Question 8. A circular park has a path of uniform width $5\ \text{m}$ running outside it. The radius of the park is $20\ \text{m}$. What is the area of the path? (Use $\pi = 3.14$)

(A) Area of park $= \pi (20)^2 = 400\pi$. Outer radius $= 20+5 = 25\ \text{m}$. Outer area $= \pi (25)^2 = 625\pi$. Area of path $= 625\pi - 400\pi = 225\pi \approx 225 \times 3.14 = 706.5\ \text{m}^2$.

(B) Area of path $= 5\pi (20+25) = 5\pi(45) = 225\pi \approx 706.5\ \text{m}^2$.

(C) Area of path $= \pi (20+5)^2 - \pi (20)^2 = \pi (25^2 - 20^2) = \pi (625 - 400) = 225\pi\ \text{m}^2$.

(D) All of the above calculations and result are correct (approximate).

Answer:

Question 9. A figure is formed by a rectangle of length $10\ \text{cm}$ and width $7\ \text{cm}$ with a semicircle attached to each of the shorter sides as diameter. What is the area of the combined figure? (Use $\pi = \frac{22}{7}$)

(A) Area of rectangle $= 10 \times 7 = 70\ \text{cm}^2$. Radius of semicircle $= 7/2 = 3.5\ \text{cm}$. Area of two semicircles $=$ Area of one circle with radius $3.5\ \text{cm} = \pi (3.5)^2 = \frac{22}{7} \times 12.25 = 38.5\ \text{cm}^2$. Total area $= 70 + 38.5 = 108.5\ \text{cm}^2$.

(B) $70 + 2 \times \frac{1}{2}\pi (7)^2 = 70 + 49\pi\ \text{cm}^2$.

(C) $70 + \pi (7/2)^2\ \text{cm}^2$.

(D) Both (A) and (C) are correct calculations/results.

Answer:

Question 10. For the figure in Question 9, what is the perimeter of the combined figure? (Use $\pi = \frac{22}{7}$)

(A) Perimeter $=$ Length $+$ Length $+$ Circumference of one circle (with radius 3.5) $= 10 + 10 + 2\pi(3.5) = 20 + 7\pi = 20 + 7 \times \frac{22}{7} = 20 + 22 = 42\ \text{cm}$.

(B) Perimeter $= 10 + 7 + 10 + 7 = 34\ \text{cm}$.

(C) Perimeter $= 20 + 2 \times \frac{1}{2}(2\pi \times 3.5) = 20 + 7\pi \approx 42\ \text{cm}$.

(D) Both (A) and (C) are correct calculations/results.

Answer:



Introduction to Solid Shapes and Basic Terms

Question 1. A child is playing with building blocks. He picks up a block that has 6 square faces, 12 edges, and 8 vertices.

Which solid shape does the block represent?

(A) Cuboid

(B) Cube

(C) Square Pyramid

(D) Triangular Prism

Answer:

Question 2. A circus tent is made of canvas. It has a circular base, a curved side tapering to a point at the top.

What standard solid shape does the main structure of the tent resemble?

(A) Cylinder

(B) Sphere

(C) Cone

(D) Pyramid

Answer:

Question 3. Look at the following objects: a book, a football, a matchbox, a die (singular of dice).

Which of these objects can be best represented by a cuboid?

(A) A book and a matchbox

(B) A football and a die

(C) A book and a die

(D) A football and a matchbox

Answer:

Question 4. A glass is typically cylindrical in shape. It has a flat base and a flat top (when full or empty) and a curved side.

How many faces, edges, and vertices does a typical open glass (ignoring thickness) have?

(A) Faces: 3, Edges: 2, Vertices: 0

(B) Faces: 2, Edges: 1, Vertices: 0

(C) Faces: 1, Edges: 1, Vertices: 1

(D) Faces: 3, Edges: 1, Vertices: 0

Answer:

Question 5. Consider Euler's formula for polyhedrons: $F + V - E = 2$. A triangular prism has 5 faces and 6 vertices.

How many edges does a triangular prism have according to Euler's formula?

(A) 6

(B) 7

(C) 8

(D) 9

Answer:

Question 6. A solid shape has a single curved surface and no edges or vertices. Every point on its surface is equidistant from a fixed point inside.

Which solid shape is being described?

(A) Cone

(B) Cylinder

(C) Sphere

(D) Hemisphere

Answer:

Question 7. A tent is made in the shape of a square pyramid. It has a square base and four triangular sides that meet at a point at the top.

How many faces, edges, and vertices does a square pyramid have?

(A) Faces: 4, Edges: 6, Vertices: 4

(B) Faces: 5, Edges: 8, Vertices: 5

(C) Faces: 5, Edges: 6, Vertices: 6

(D) Faces: 4, Edges: 4, Vertices: 4

Answer:

Question 8. A solid shape has two identical parallel polygonal bases and its lateral faces are rectangles or parallelograms connecting corresponding sides of the bases.

Which type of solid shape is this?

(A) Pyramid

(B) Prism

(C) Cone

(D) Cylinder

Answer:

Question 9. A solid shape has a circular base and a curved surface that tapers uniformly to a single vertex (apex) directly above the center of the base.

Which standard solid is being described?

(A) Cylinder

(B) Sphere

(C) Cone

(D) Hemisphere

Answer:

Question 10. Applying Euler's formula to a triangular pyramid (tetrahedron). A triangular pyramid has 4 faces (4 triangles), 6 edges, and 4 vertices.

Does Euler's formula hold for a triangular pyramid? ($F + V - E = 2$)

(A) Yes, $4+4-6 = 2$.

(B) Yes, $4+6-4 = 6$.

(C) No, $4+4-6 \neq 2$.

(D) Euler's formula only applies to prisms, not pyramids.

Answer:



Surface Area of Standard Solids

Question 1. A gift box is in the shape of a cube with a side length of $15\ \text{cm}$. The box needs to be wrapped with gift paper on all its outer surfaces.

What is the total surface area of the box that needs to be covered?

(A) $225\ \text{cm}^2$

(B) $900\ \text{cm}^2$

(C) $1350\ \text{cm}^2$

(D) $3375\ \text{cm}^3$

Answer:

Question 2. A rectangular room is $6\ \text{m}$ long, $4\ \text{m}$ wide, and $3\ \text{m}$ high. The walls and ceiling need to be painted. The area of the floor should not be painted.

What is the total area to be painted?

(A) $2(6 \times 4 + 4 \times 3 + 3 \times 6) = 2(24+12+18) = 2(54) = 108\ \text{m}^2$ (TSA)

(B) $2(6+4) \times 3 + 6 \times 4 = 2(10) \times 3 + 24 = 60 + 24 = 84\ \text{m}^2$ (LSA + Area of ceiling)

(C) $6 \times 4 \times 3 = 72\ \text{m}^3$ (Volume)

(D) $2(6+4) \times 3 = 60\ \text{m}^2$ (LSA)

Answer:

Question 3. A cylindrical pillar has a base radius of $35\ \text{cm}$ and a height of $5\ \text{m}$. It needs to be painted on its curved surface.

What is the curved surface area to be painted? (Use $\pi = \frac{22}{7}$ and convert units to metres).

(A) $2\pi (0.35)(5) = 3.5\pi\ \text{m}^2 \approx 11\ \text{m}^2$

(B) $2\pi (35)(500)\ \text{cm}^2$

(C) $2\pi (35)(5)\ \text{cm}^2$

(D) $\pi (0.35)^2 (5)\ \text{m}^3$

Answer:

Question 4. A cone-shaped cap has a base radius of $7\ \text{cm}$ and a slant height of $25\ \text{cm}$. The material needed to make the cap (excluding the base) is its curved surface area.

What is the curved surface area of the cap? (Use $\pi = \frac{22}{7}$)

(A) $175\pi\ \text{cm}^2$

(B) $550\ \text{cm}^2$

(C) $616\ \text{cm}^2$

(D) $1100\ \text{cm}^2$

Answer:

Question 5. A spherical ball has a radius of $10.5\ \text{cm}$. What is its surface area? (Use $\pi = \frac{22}{7}$)

(A) $1386\ \text{cm}^2$

(B) $5544\ \text{cm}^2$

(C) $10.5 \times 2\pi = 21\pi\ \text{cm}$

(D) $4\pi (10.5)^3\ \text{cm}^3$

Answer:

Question 6. A hemispherical bowl is made of steel. The inner radius is $5\ \text{cm}$. The area of the inner curved surface needs to be painted.

What is the area of the inner curved surface to be painted? (Use $\pi = 3.14$)

(A) $2\pi (5)^2 = 50\pi \approx 157\ \text{cm}^2$

(B) $\pi (5)^2 = 25\pi \approx 78.5\ \text{cm}^2$

(C) $3\pi (5)^2 = 75\pi \approx 235.5\ \text{cm}^2$

(D) $4\pi (5)^2 = 100\pi \approx 314\ \text{cm}^2$

Answer:

Question 7. A solid wooden toy is in the shape of a cone mounted on a hemisphere. The radius of the base of the cone is $3.5\ \text{cm}$ and its height is $12\ \text{cm}$. The toy is to be painted.

To find the total area to be painted, you need to calculate the curved surface area of the cone and the curved surface area of the hemisphere. First, find the slant height of the cone.

(A) $\sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5\ \text{cm}$

(B) $\sqrt{12^2 - 3.5^2} = \sqrt{144 - 12.25} = \sqrt{131.75} \approx 11.48\ \text{cm}$

(C) $12 + 3.5 = 15.5\ \text{cm}$

(D) $12.5\ \text{cm}$

Answer:

Question 8. For the wooden toy in Question 7 (cone on hemisphere, $r=3.5\ \text{cm}$, $h=12\ \text{cm}$, $l=12.5\ \text{cm}$), what is the curved surface area of the cone? (Use $\pi = \frac{22}{7}$)

(A) $\pi r l = \frac{22}{7} \times 3.5 \times 12.5 = 11 \times 12.5 = 137.5\ \text{cm}^2$

(B) $\pi r^2 = \frac{22}{7} \times (3.5)^2 = 38.5\ \text{cm}^2$

(C) $2\pi r h = 2\pi (3.5)(12) = 84\pi \approx 264\ \text{cm}^2$

(D) $137.5\ \text{cm}^2$

Answer:

Question 9. For the wooden toy in Question 7 (cone on hemisphere, $r=3.5\ \text{cm}$, $h=12\ \text{cm}$, $l=12.5\ \text{cm}$), what is the curved surface area of the hemisphere? (Use $\pi = \frac{22}{7}$)

(A) $\pi r^2 = 38.5\ \text{cm}^2$

(B) $2\pi r^2 = 2 \times 38.5 = 77\ \text{cm}^2$

(C) $3\pi r^2 = 3 \times 38.5 = 115.5\ \text{cm}^2$

(D) $4\pi r^2 = 4 \times 38.5 = 154\ \text{cm}^2$

Answer:

Question 10. For the wooden toy in Question 7 (cone on hemisphere), what is the total surface area to be painted?

(A) CSA of cone $+$ CSA of hemisphere $= 137.5 + 77 = 214.5\ \text{cm}^2$

(B) TSA of cone $+$ TSA of hemisphere

(C) CSA of cone $+$ CSA of hemisphere $+$ Area of common base

(D) $214.5\ \text{cm}^2$

Answer:



Volume of Standard Solids

Question 1. A storage box is in the shape of a cuboid. Its dimensions are $1.2\ \text{m} \times 0.8\ \text{m} \times 0.5\ \text{m}$. What is the volume of the box?

(A) $0.48\ \text{m}^3$

(B) $4.8\ \text{m}^3$

(C) $480\ \text{m}^3$

(D) $2.5\ \text{m}$

Answer:

Question 2. A cylindrical water tank has a base radius of $2.1\ \text{m}$ and a height of $5\ \text{m}$. What is the capacity of the tank in cubic metres? (Use $\pi = \frac{22}{7}$)

(A) $\pi (2.1)^2 \times 5 = \frac{22}{7} \times 4.41 \times 5 = 22 \times 0.63 \times 5 = 22 \times 3.15 = 69.3\ \text{m}^3$

(B) $693\ \text{m}^3$

(C) $138.6\ \text{m}^3$

(D) $69.3\ \text{m}^3$

Answer:

Question 3. How many litres of water can the tank in Question 2 hold? ($1\ \text{m}^3 = 1000\ \text{L}$)

(A) $69.3\ \text{L}$

(B) $693\ \text{L}$

(C) $6930\ \text{L}$

(D) $69300\ \text{L}$

Answer:

Question 4. An ice cream cone has a radius of $3\ \text{cm}$ and a height of $10\ \text{cm}$. Assuming it's a perfect cone (without the ice cream), what is its volume? (Use $\pi = 3.14$)

(A) $\frac{1}{3}\pi (3)^2 \times 10 = \frac{1}{3}\pi \times 9 \times 10 = 30\pi \approx 30 \times 3.14 = 94.2\ \text{cm}^3$

(B) $942\ \text{cm}^3$

(C) $30\pi\ \text{cm}^3$

(D) Both (A) and (C) are correct (approximate/exact).

Answer:

Question 5. A spherical ball of diameter $18\ \text{cm}$ is made of solid rubber. What is the volume of rubber in the ball? (Use $\pi = 3.14$)

(A) Radius $= 9\ \text{cm}$. Volume $= \frac{4}{3}\pi (9)^3 = \frac{4}{3}\pi \times 729 = 4\pi \times 243 = 972\pi \approx 972 \times 3.14 = 3052.08\ \text{cm}^3$.

(B) $\frac{4}{3}\pi (18)^3$

(C) $4\pi (9)^2$

(D) $3052.08\ \text{cm}^3$

Answer:

Question 6. A hemispherical bowl has an inner radius of $14\ \text{cm}$. What is the volume of water it can hold when completely filled? (Use $\pi = \frac{22}{7}$)

(A) $\frac{2}{3}\pi (14)^3 = \frac{2}{3} \times \frac{22}{7} \times 2744 = \frac{44}{21} \times 2744 = \frac{44 \times 392}{3} = 17248/3 \approx 5749.33\ \text{cm}^3$

(B) $5749.33\ \text{cm}^3$

(C) $\frac{4}{3}\pi (14)^3$

(D) $2\pi (14)^2$

Answer:

Question 7. A solid metallic cuboid with dimensions $11\ \text{cm} \times 10\ \text{cm} \times 5\ \text{cm}$ is melted and recast into a solid cylinder of height $10\ \text{cm}$. What is the radius of the cylinder?

(A) $3\ \text{cm}$

(B) $3.5\ \text{cm}$

(C) $7\ \text{cm}$

(D) $14\ \text{cm}$

Answer:

Question 8. A spherical sweet (ladoo) has a radius of $4\ \text{cm}$. If it is melted and reshaped into a smaller spherical sweet of radius $2\ \text{cm}$, how many smaller sweets can be made?

(A) 2

(B) 4

(C) 8

(D) 16

Answer:

Question 9. A cylindrical pipe has an inner diameter of $5\ \text{cm}$ and an outer diameter of $6\ \text{cm}$. It is $20\ \text{cm}$ long. What is the volume of metal used in the pipe? (Use $\pi = 3.14$)

(A) Volume $= \pi (3^2 - 2.5^2) \times 20 = \pi (9 - 6.25) \times 20 = \pi (2.75) \times 20 = 55\pi \approx 55 \times 3.14 = 172.7\ \text{cm}^3$.

(B) $110\pi\ \text{cm}^3$

(C) $172.7\ \text{cm}^3$

(D) Both (A) and (C) are correct (approximate/exact).

Answer:

Question 10. A conical pile of wheat has a base diameter of $14\ \text{m}$ and a height of $3\ \text{m}$. What is the volume of wheat in the pile? (Use $\pi = \frac{22}{7}$)

(A) $\frac{1}{3}\pi (7)^2 \times 3 = \pi \times 49 = \frac{22}{7} \times 49 = 22 \times 7 = 154\ \text{m}^3$

(B) $154\ \text{m}^3$

(C) $\frac{1}{3}\pi (14)^2 \times 3 = 196\pi\ \text{m}^3$

(D) $462\ \text{m}^3$

Answer:



Surface Area and Volume of Combined Solids

Question 1. A solid is in the shape of a cone standing on a hemisphere, both having a radius equal to $1\ \text{cm}$. The height of the cone is also $1\ \text{cm}$. What is the volume of the solid? (Use $\pi = 3.14$)

(A) Volume of cone $= \frac{1}{3}\pi (1)^2 (1) = \frac{1}{3}\pi\ \text{cm}^3$. Volume of hemisphere $= \frac{2}{3}\pi (1)^3 = \frac{2}{3}\pi\ \text{cm}^3$. Total Volume $= \frac{1}{3}\pi + \frac{2}{3}\pi = \pi\ \text{cm}^3 \approx 3.14\ \text{cm}^3$.

(B) $\frac{1}{3}\pi\ \text{cm}^3$

(C) $\frac{2}{3}\pi\ \text{cm}^3$

(D) $\pi\ \text{cm}^3$

Answer:

Question 2. A tent is in the shape of a cylinder surmounted by a cone. The height and diameter of the cylindrical part are $2.1\ \text{m}$ and $4\ \text{m}$ respectively, and the slant height of the conical part is $2.8\ \text{m}$. The volume of air inside the tent is needed.

What is the volume of the cylindrical part? (Use $\pi = \frac{22}{7}$)

(A) Radius $= 2\ \text{m}$. Volume $= \pi (2)^2 (2.1) = 4\pi \times 2.1 = 8.4\pi = 8.4 \times \frac{22}{7} = 1.2 \times 22 = 26.4\ \text{m}^3$.

(B) $264\ \text{m}^3$

(C) $10.5\pi\ \text{m}^3$

(D) $5.88\pi\ \text{m}^3$

Answer:

Question 3. For the tent in Question 2, find the height of the conical part.

(A) $h = \sqrt{l^2 - r^2} = \sqrt{2.8^2 - 2^2} = \sqrt{7.84 - 4} = \sqrt{3.84} \approx 1.96\ \text{m}$.

(B) $2.1\ \text{m}$

(C) $2.8\ \text{m}$

(D) $1.96\ \text{m}$ (approx)

Answer:

Question 4. For the tent in Question 2, what is the volume of the conical part? (Use $\pi = \frac{22}{7}$, $h \approx 1.96$ from Q3)

(A) $\frac{1}{3}\pi (2)^2 (2.1) = \frac{1}{3}\pi (8.4) = 2.8\pi \approx 8.8\ \text{m}^3$.

(B) $\frac{1}{3}\pi (2)^2 (1.96) \approx \frac{4}{3}\pi (1.96) \approx 8.2\ \text{m}^3$.

(C) $26.4\ \text{m}^3$

(D) $8.2\ \text{m}^3$ (approx)

Answer:

Question 5. For the tent in Question 2, what is the total volume of air inside the tent?

(A) Volume of cylinder $+$ Volume of cone $\approx 26.4 + 8.2 = 34.6\ \text{m}^3$ (approx)

(B) $26.4\ \text{m}^3$

(C) $8.2\ \text{m}^3$

(D) $34.6\ \text{m}^3$ (approx)

Answer:

Question 6. A solid is in the shape of a hemisphere mounted by a cylinder and surmounted by a cone. The common radius is $7\ \text{cm}$. The height of the cylinder is $6\ \text{cm}$ and the height of the cone is $4\ \text{cm}$. What is the total volume of the solid? (Use $\pi = \frac{22}{7}$)

(A) Volume $=$ Volume of hemisphere $+$ Volume of cylinder $+$ Volume of cone $= \frac{2}{3}\pi (7)^3 + \pi (7)^2 (6) + \frac{1}{3}\pi (7)^2 (4)$

(B) $= \pi (7)^2 (\frac{2}{3} \times 7 + 6 + \frac{1}{3} \times 4) = 49\pi (\frac{14}{3} + 6 + \frac{4}{3}) = 49\pi (\frac{18}{3} + 6) = 49\pi (6+6) = 49\pi (12) = 588\pi\ \text{cm}^3$

(C) $588\pi = 588 \times \frac{22}{7} = 84 \times 22 = 1848\ \text{cm}^3$.

(D) All of the above calculations and result are correct.

Answer:

Question 7. A cylindrical vessel with internal diameter $10\ \text{cm}$ and height $10.5\ \text{cm}$ is full of water. A solid in the shape of a cone of base diameter $7\ \text{cm}$ and height $6\ \text{cm}$ is placed in it. The volume of water left in the cylinder is required.

What is the volume of the cylinder? (Use $\pi = \frac{22}{7}$)

(A) Radius $= 5\ \text{cm}$. Volume $= \pi (5)^2 (10.5) = 25\pi \times 10.5 = 262.5\pi = 262.5 \times \frac{22}{7} = 37.5 \times 22 = 825\ \text{cm}^3$.

(B) $825\ \text{cm}^3$

(C) $1650\ \text{cm}^3$

(D) $262.5\pi\ \text{cm}^3$

Answer:

Question 8. For the scenario in Question 7, what is the volume of the cone? (Use $\pi = \frac{22}{7}$)

(A) Radius $= 3.5\ \text{cm}$. Volume $= \frac{1}{3}\pi (3.5)^2 (6) = \frac{1}{3}\pi (12.25)(6) = \pi (12.25)(2) = 24.5\pi = 24.5 \times \frac{22}{7} = 3.5 \times 22 = 77\ \text{cm}^3$.

(B) $77\ \text{cm}^3$

(C) $154\ \text{cm}^3$

(D) $24.5\pi\ \text{cm}^3$

Answer:

Question 9. For the scenario in Question 7, what is the volume of water left in the cylinder after placing the cone?

(A) Volume of cylinder $-$ Volume of cone $= 825 - 77 = 748\ \text{cm}^3$.

(B) $748\ \text{cm}^3$

(C) Volume of cone

(D) Volume of cylinder

Answer:

Question 10. A wooden article is made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is $10\ \text{cm}$ and its base radius is $3.5\ \text{cm}$, what is the volume of wood in the article? (Use $\pi = \frac{22}{7}$)

(A) Volume of cylinder $-$ Volume of two hemispheres $= \pi (3.5)^2 (10) - 2 \times \frac{2}{3}\pi (3.5)^3 = 38.5 \times 10 - \frac{4}{3}\pi (12.25 \times 3.5) = 385 - \frac{4}{3}\pi (42.875) = 385 - \frac{4}{3} \times \frac{22}{7} \times 42.875 \approx 385 - 179.67 = 205.33\ \text{cm}^3$.

(B) Volume of cylinder $+$ Volume of two hemispheres

(C) $385 - 2 \times (\frac{2}{3} \times \frac{22}{7} \times (3.5)^3) = 385 - \frac{4}{3} \times 38.5 \times 3.5 = 385 - \frac{154}{3} \times 3.5 = 385 - \frac{539}{3} \approx 385 - 179.67 = 205.33\ \text{cm}^3$.

(D) Both (A) and (C) are correct calculations/approximate results.

Answer:



Conversion of Solids and Frustum of a Cone

Question 1. A solid metallic sphere of radius $9\ \text{cm}$ is melted and recast into a solid cylinder of radius $6\ \text{cm}$. What is the height of the cylinder?

(A) Volume of sphere $= \frac{4}{3}\pi (9)^3 = \frac{4}{3}\pi \times 729 = 4\pi \times 243 = 972\pi\ \text{cm}^3$. Volume of cylinder $= \pi (6)^2 h = 36\pi h$. Equating volumes: $972\pi = 36\pi h \implies h = 972/36 = 27\ \text{cm}$.

(B) $18\ \text{cm}$

(C) $27\ \text{cm}$

(D) $36\ \text{cm}$

Answer:

Question 2. A $14\ \text{m}$ deep well with inner diameter $6\ \text{m}$ is dug. The earth taken out is evenly spread all around the well to form an embankment $4\ \text{m}$ wide. What is the height of the embankment? (Use $\pi = \frac{22}{7}$)

(A) Volume of earth $=$ Volume of cylinder (well) $= \pi (3)^2 (14) = 9\pi \times 14 = 126\pi\ \text{m}^3$. Area of embankment ring $= \pi (R_{outer}^2 - R_{inner}^2)$. Inner radius of embankment $=$ radius of well $= 3\ \text{m}$. Outer radius of embankment $= 3+4 = 7\ \text{m}$. Area of embankment ring $= \pi (7^2 - 3^2) = \pi (49 - 9) = 40\pi\ \text{m}^2$. Volume of embankment $=$ Area of ring $\times$ height $= 40\pi \times h$. Equating volumes: $126\pi = 40\pi h \implies h = 126/40 = 63/20 = 3.15\ \text{m}$.

(B) $4.2\ \text{m}$

(C) $3.15\ \text{m}$

(D) $2.8\ \text{m}$

Answer:

Question 3. A frustum of a cone has radii of $10\ \text{cm}$ and $4\ \text{cm}$ and height $8\ \text{cm}$. What is its slant height?

(A) $l = \sqrt{8^2 + (10-4)^2} = \sqrt{64 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10\ \text{cm}$.

(B) $8\ \text{cm}$

(C) $12\ \text{cm}$

(D) $14\ \text{cm}$

Answer:

Question 4. For the frustum in Question 3 (radii $10\ \text{cm}$, $4\ \text{cm}$, height $8\ \text{cm}$, slant height $10\ \text{cm}$), what is its curved surface area? (Use $\pi = 3.14$)

(A) $\pi (10+4) \times 10 = 140\pi \approx 140 \times 3.14 = 439.6\ \text{cm}^2$.

(B) $439.6\ \text{cm}^2$

(C) $140\pi\ \text{cm}^2$

(D) Both (A) and (C) are correct (approximate/exact).

Answer:

Question 5. For the frustum in Question 3 (radii $10\ \text{cm}$, $4\ \text{cm}$, height $8\ \text{cm}$), what is its volume? (Use $\pi = 3.14$)

(A) $V = \frac{1}{3}\pi (8) (10^2 + 4^2 + 10 \times 4) = \frac{8}{3}\pi (100 + 16 + 40) = \frac{8}{3}\pi (156) = 8 \times 52\pi = 416\pi \approx 416 \times 3.14 = 1306.24\ \text{cm}^3$.

(B) $1306.24\ \text{cm}^3$

(C) $416\pi\ \text{cm}^3$

(D) All of the above calculations and result are correct (approximate/exact).

Answer:

Question 6. A solid metallic cone of height $20\ \text{cm}$ and radius $5\ \text{cm}$ is melted and recast into several spherical balls of radius $1\ \text{cm}$. How many spherical balls can be made? (Use $\pi$ as $\pi$)

(A) Volume of cone $= \frac{1}{3}\pi (5)^2 (20) = \frac{500\pi}{3}$. Volume of one ball $= \frac{4}{3}\pi (1)^3 = \frac{4\pi}{3}$. Number of balls $= \frac{\text{Volume of cone}}{\text{Volume of one ball}} = \frac{500\pi/3}{4\pi/3} = \frac{500}{4} = 125$.

(B) 125

(C) 25

(D) Both (A) and (B) are correct calculations/result.

Answer:

Question 7. A container is in the shape of a frustum of a cone, open at the top, with height $16\ \text{cm}$ and radii of its lower and upper circular ends as $8\ \text{cm}$ and $20\ \text{cm}$ respectively. The volume of milk which can completely fill the container is required. (Use $\pi = 3.14$)

(A) $V = \frac{1}{3}\pi (16) (8^2 + 20^2 + 8 \times 20) = \frac{16}{3}\pi (64 + 400 + 160) = \frac{16}{3}\pi (624) = 16\pi (208) = 3328\pi \approx 3328 \times 3.14 = 10449.92\ \text{cm}^3$.

(B) $10449.92\ \text{cm}^3$ (approx)

(C) $3328\pi\ \text{cm}^3$

(D) All of the above calculations and result are correct (approximate/exact).

Answer:

Question 8. How many litres of milk can the container in Question 7 hold? ($1\ \text{L} = 1000\ \text{cm}^3$).

(A) $10.44992\ \text{L}$ (approx)

(B) $104.5\ \text{L}$ (approx)

(C) $3.328\pi\ \text{L}$ (exact)

(D) Both (A) and (C) are correct (approximate/exact).

Answer:

Question 9. A cone of height $H$ is cut by a plane parallel to the base at a height $h'$ from the vertex. If the radius of the original base is $R$, what is the radius $r'$ of the smaller cone cut off at the top?

(A) $r' = R$

(B) $r' = R \frac{h'}{H}$

(C) $r' = R \frac{H}{h'}$

(D) $r' = R - h'$

Answer:

Question 10. If the cone in Question 9 is cut at half its height from the base, meaning $h' = H/2$ (height of small cone from vertex is $H/2$), what is the ratio of the volume of the small cone to the volume of the original cone?

(A) $1:2$

(B) $1:4$

(C) $1:8$

(D) $1:16$

Answer: